(i) By Lemma REF , we have the \(E_1^{*,*}\) -term as desired, and by definition we obtain \(d_1=j_1k_1:E_1^{p,q}\rightarrow E_1^{p+1,q}\) . We only need to consider the case when \(q\)  is even,
otherwise \(d_1=0.\)  By (REF ) for odd \(p\)  (the case when \(p\)  is
even, is similar), we have a large commutative diagram
\({& &\vdots &\vdots &\vdots & &\\& 0[r]& K_{p+1}^e[u]_D[r]^{i}& K_{p}^e[u]_D[r]^{j}&0 [u]_D[r]&0\\&0[r]&K_{p+1}^o[u]_D[r]^i& K_{p}^o[u]_{D}[r]^{j}&\Omega ^p(M) [u]_D[r]&0\\&0[r]& K_{p+1}^e[u]_D[r]^{i}& K_{p}^e[u]_D[r]^{j}&0 [u]_D[r]&0\\& &\vdots [u]_D &\vdots [u]_D &\vdots [u]_D & &\\}\) 
where the rows are exact and the columns are cochain complexes.
Let \(x_p\in \Omega ^p(M)\cong H_D^{p+q}(K_p/K_{p+1})\cong E_1^{p,q}\) 
and
\(x=\sum _{i=0}^{[\frac{n-p}{2}]}x_{p+2i}\) 
be an (inhomogeneous) form, where \(x_{p+2i}\)  is a \((p+2i)\) -form (\(0\le i\le [\frac{n-p}{2}]\) ). Then \(x\in K_p^o\) , \(jx=x_p\)  and \(Dx\in K_p^e\) . Also \(Dx\in K_{p+1}^e\) . By the definition of the
homomorphism \(\delta \)  in (REF ), we have
\(k_1x_p=[Dx]_D,\) 
where \([\ \ ]_D\)  is the
cohomology class in \(H_D^{\ast }(K_{p+1})\) . The class \([Dx]_D\)  is
well defined and independent of the choices of \(x_{p+2i}\)  \((1 \le i\le [\frac{n-p}{2}])\)  (see [1]}).
Choose \(x_{p+2i}=0\)  (\(1 \le i\le [\frac{n-p}{2}]\) ). Then we have
\(k_1x_p & = & [Dx]_D\\&=&[dx_p+H\wedge x_p]_D\\& = & [dx_p + \sum _{l=1}^{[\frac{n-1}{2}]}H_{2l+1}\wedge x_p]_D \in H_D^{p+q+1}(K_{p+1}).\) 
Thus one obtains
\(d_1x_p &=&(j_1 k_1)x_p\\&=& j_1(k_1(x_p))\\& = & j_1[dx_p + \sum _{l=1}^{[\frac{n-1}{2}]}H_{2l+1}\wedge x_p]_D\\& = & dx_p.\) 
(ii) By the definition of the spectral sequence and (i), one obtains
that \(E_2^{p,q}\cong H^{p}(M)\)  when \(q\)  is even, and \(E_2^{p,q}=0\) 
when \(q\)  is odd. Note that \(d_2: E^{p, q}_2 \rightarrow E_2^{p+2, q-1}\) . It
follows that \(d_2=0\)  by degree reasons.
(iii) Note that \([x_p]_3\in E_3^{p, q}\)  implies \(dx_p=0\) . Choose
\(x_{p+2i}=0\)  for \(1 \le i\le [\frac{n-p}{2}]\) , and we get
\([Dx]_D=[H\wedge x_p]_D =[ \sum _{l=1}^{[\frac{n-1}{2}]}H_{2l+1}\wedge x_p]_D\in H_D^{p+q+1}(K_{p+1}),\) 
where \(x\)  is given in the proof of (i). Note
that
\({&H_D^{p+q+1}(K_{p+1})&[l]_{i_1^2} H_D^{p+q+1}(K_{p+3})[r]^{j_1}&H_D^{p+q+1}(K_{p+3}/K_{p+4})\\&[Dx]_D[r]^{(i_1^{-1})^2}&[Dx]_D[r]^{j_1}& H_3\wedge x_p. \\}\) 
It follows that
\(\begin{array}{lll}d_3[x_p]_3 & =& j_3k_3[x_p]_3\\\end{array}&=&j_3(k_1x_p)\\\) 
 = j3[Dx]D
= [j1((i1-1)2[Dx]D)]3
= [H3xp]3,


where the first,
second and fourth identities follow from the definitions of \(d_3\) ,
\(k_3\)  and \(j_3\)  respectively, and the third and the last identities
follow from (REF ) and (REF ). By (ii), \(d_2=0\) , so
\(E_3^{p,q}=E_2^{p,q}.\)  Then we have
\(d_3[x_p]=[H_3\wedge x_p].\) 
